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Q13E

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Found in: Page 443

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Determine whether the series is convergent or divergent. If its convergent, find its sum.$$\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{(n - 1)}}}}}$$

The given geometric series is divergent.

See the step by step solution

## Rewriting the equation.

\begin{aligned}{l}\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {{3^n}{e^{{{(n - 1)}^{ - 1}}}}} \\\end{aligned}

$$= \sum\limits_{n = 1}^\infty {{3^n}{e^{(1 - n)}}}$$

## Writing in terms of b and q.

$$\sum\limits_{n = 1}^\infty {{3^n}{e^{(1 - n)}}}$$ is a geometric series with first term b = $$\frac{3}{e}$$and common ratio q = $$\frac{3}{e}$$.

## Applying ratio test.

Since $$\left| q \right| = \left| {\frac{3}{e}} \right| > 1$$, then by ratio test the series is divergent.

Hence, $$\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{n - 1}}}}}$$ is divergent.